Question

15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remaining nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus had a mass of 222 u, what speed does the alpha particle have when it is emitted?
Use conservation of momentum to solve.

Answer 1

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

$p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =$  

where:  

$M =222u$ is the mass of the original nucleus

$v=420 m/s$ is the initial velocity of the nucleus

$m_1 = 4 u$ is the mass of the alpha particle

$v_1$ is the final velocity of the alpha particle

$m_2 = 222u-4u = 218 u$ is the mass of the daughter nucleus

$v_2 = 350 m/s$ is the final velocity of the nucleus

Solving for $v_1$, we  find the final velocity of the alpha particle:

$v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s$

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